I've seen a few arguments for why that is the case, but I was wondering if there was anything wrong with my argument.
One property of the outer measure $m_*$ is that if $E \subset \mathbb{R}^d$, then $m_*(E) = \inf m_*(O)$, where the infimum is taken over all open sets such that $E \subset O$.
Now, since the irrationals are dense in $\mathbb{R}$, then they are dense in $[a,b]$, and so, the smallest open set containing the irrationals is $(a,b)$. Then, it must be that:
$$ m_*(\text{irrationals in $[a,b]$}) = m_*((a,b)) = b-a $$
I'm wondering if there are any flaws in the reasoning?
Yes, it is wrong. Take a rational $q\in(a,b)$. Then $[a,b]\setminus\{q\}$ is an open subset of $[a,b]$ containing the irrationals of $[a,b]$.